直接就是板子了，但是呢，我去了個重。

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<string>
#include<set>
#include<algorithm>
#include<vector>
#include<queue>
#include<list>
#include<cmath>
#include<cstring>
#include<map>
#include
 
<stack>
using namespace std;
#define INF 0x3f3f3f3f
#define maxn 1005
#define ull unsigned long long
#define ll long long
#define hashmod 99999839
#define mod 9997
struct func{
    ll p;
    ll y;
}a[maxn];
int n;
ll m[maxn],y[maxn];
bool cmp(const func& a,const func& b){
    
 
if(a.p != b.p) return a.p < b.p;
    return a.y < b.y;
}
ll excgcd(ll a,ll b,ll& x,ll& y){
    if(!b){x = 1,y = 0;return a;}
    ll g = excgcd(b,a % b,x,y);
    ll t = x;
    x = y;
    y = t - a/b * y;
    return g;
}
void check(){
    sort(a + 1,a + n + 1,cmp);
    int j = 2
 
;
    for(int i = 2;i <= n;++i){
        if(a[i].p == a[i-1].p && a[i].y == a[i-1].y) continue;
        a[j] = a[i];
        ++j;
    }
    n = j - 1;
}
void solve(){
    ll M = 1,x = 0,y = 0,ans = 0;//[m_1,m_2,...,m_n]
    for(int i = 1;i <= n;++i) M *= a[i].p;
    for(int i = 1;i <= n;++i){
        m[i] = M / a[i].p;
        ll g = excgcd(m[i],a[i].p,x,y);
        x = x + (abs(x)/(a[i].p/g) + 1) * (a[i].p/g);
        x = x % (a[i].p/g);//將x擴展為正數
        ans = ans + x * m[i] * a[i].y;
        if(ans >= M) ans %= M;
    }
    printf("%lld\n",ans);
}
int main(){
//    freopen("a.in","r",stdin);
//    freopen("b.out","w",stdout);
    while(~scanf("%d",&n)){
        for(int i = 1;i <= n;++i) scanf("%lld%lld",&a[i].p,&a[i].y);
        check();
        solve();
    }
    return 0;
}

51Nod - 1079 中國剩余定理